Let M be the mass and L be the length of a th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The moment of inertia $(I)$ of a thin uniform rod rotating about an axis passing through its center and perpendicular to its length is given by $I

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) The moment of inertia $(I)$ of a thin uniform rod rotating about an axis passing through its center and perpendicular to its length is given by $I_1 = \frac{ML^2}{12}$.Since $I = MK^2$,where $K$ is the radius of gyration,we have $MK_1^2 = \frac{ML^2}{12}$,which gives $K_1 = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$.In the second case,the axis of rotation passes through one end and is perpendicular to the length of the rod. The moment of inertia is $I_2 = \frac{ML^2}{3}$.Using $MK_2^2 = \frac{ML^2}{3}$,we get $K_2 = \frac{L}{\sqrt{3}}$.The ratio of the radius of gyration in the first case to the second case is $\frac{K_1}{K_2} = \frac{L / (2\sqrt{3})}{L / \sqrt{3}} = \frac{1}{2}$.

Similar Questions

Find the moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane.

The radius of gyration of a body depends upon

If a spherical planet of mass $M$ and radius $R$ suddenly shrinks to half its size,and its mass reduces to half,what is the new moment of inertia of the planet about its diameter?

Vedclass Test Series

Exam Paper Generator

Online Exam Module